$Q_2=600 J$

            ${T_1} = {127^ \circ }C = 400\,K$

            ${T_2} = ?$

            $\eta  = ?$

Efficiency of carnot engine,

$\eta  = \frac{W}{Q_1} \times 100\% $

$or,\,\,\eta  = \frac{{{Q_2} - {Q_1}}}{{{Q_1}}} \times 100\% $

$or,\,\,\,\eta  = \frac{{1000 - 600}}{{1000}} \times 100\% $

$\eta  = 40\% $

$Now,for\,carnot\,cycle\frac{{{Q_2}}}{{{Q_1}}} = \frac{{{T_2}}}{{{T_1}}}$

$\frac{{600}}{{1000}} = \frac{{{T_2}}}{{400}}$

${T_2} = \frac{{600 \times 400}}{{1000}} = 240\,K = 240 - 273$

$\therefore {T_2} =  - {33^ \circ }C$